A ball weighing $10 g m$ hits a hard surface vertically with a speed of $5 {m s}^{- 1}$ and rebounds with the same speed. The ball remains in contact with the surface for $0.01 s$ . The average force exerted by the surface on ball is :

100 N

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10 N

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1 N

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0.1 N

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Solution

The correct option is B $10 N$
Given,

$v = 5 m / s$

$m = 10 g = 0.01 k g$

$t = 0.01 s$

Newton 2nd law,

$F = \frac{Δ P}{t}$

$F = \frac{m v - (- m v)}{t} = \frac{2 m v}{t}$

$F = \frac{2 \times 0.01 \times 5}{0.01} = 10 N$

$F = 10 N$

The correct option is B.

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A ball weighing 10 gm hits a hard surface vertically with a speed of 5 ms−1 and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on ball is :

The correct option is B 10 NGiven,v=5m/sm=10g=0.01kgt=0.01sNewton 2nd law,F=ΔPtF=mv−(−mv)t=2mvtF=2×0.01×50.01=10NF=10NThe correct option is B.

A ball weighing $10 g m$ hits a hard surface vertically with a speed of $5 {m s}^{- 1}$ and rebounds with the same speed. The ball remains in contact with the surface for $0.01 s$ . The average force exerted by the surface on ball is :

The correct option is B $10 N$
Given,

$v = 5 m / s$

$m = 10 g = 0.01 k g$

$t = 0.01 s$

Newton 2nd law,

$F = \frac{Δ P}{t}$

$F = \frac{m v - (- m v)}{t} = \frac{2 m v}{t}$

$F = \frac{2 \times 0.01 \times 5}{0.01} = 10 N$

$F = 10 N$

The correct option is B.